# Stokes' first problem
> [!link] Related lecture material
> This derivation is part of [[2-What-is-Viscosity|Lecture 2: Viscosity and Momentum Diffusion]]
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## Physical Setup
A semi-infinite fluid initially at rest. At $t=0$, a plate at $y=0$ suddenly moves with velocity $u_0$.
Initial and Boundary Conditions:
$\begin{align}
\vec{u}(x,y,t=0) &= 0 \\
\text{At } t = 0^{+}, \quad u(y=0) &= u_0
\end{align}$
---
## Simplifying Assumptions
We assume spatial uniformity in the $x$-direction:
$\boxed{\frac{\partial()}{\partial x} = 0}$
Ignoring any initial transients, this leads to:
From continuity equation:
$\frac{\partial v}{\partial y} = 0$
Combined with no-slip at the plate:
$v(y=0) = 0 \quad \Rightarrow \quad v = 0 \quad \text{for all } y$
---
## Governing Equations
$x$-momentum equation:
$\rho \frac{\partial u}{\partial t} = - \frac{\partial p}{\partial x} + \mu \frac{\partial^2 u}{\partial y^2}$
$y$-momentum equation:
$\frac{\partial p}{\partial y} = 0$
Pressure gradient: For an open container with $(p,u,v) = 0$ far from the plate:
$\frac{\partial p}{\partial x} = 0$
Simplified momentum equation:
$\boxed{\rho \frac{\partial u}{\partial t} = \mu \frac{\partial^2 u}{\partial y^2}}$
or equivalently:
$\boxed{\frac{\partial u}{\partial t} = \nu \frac{\partial^2 u}{\partial y^2}}$
where $\nu = \mu/\rho$ is the kinematic viscosity.
> [!note] Key Insight
> We cannot directly integrate this PDE as $u = f(y,t)$. We need a more sophisticated approach.
---
## Boundary Conditions
$\begin{align}
u(y,0) &= 0 \quad \text{(initially at rest)} \\
u(0,t) &= u_0 \quad \text{(plate velocity)} \\
u(\infty,t) &= 0 \quad \text{(far-field BC)}
\end{align}$
---
## Similarity Solution Approach
Using dimensional analysis, we seek a solution of the form:
$u = \phi\left(u_0, y, t, \frac{\mu}{\rho} = \nu\right)$
By dimensional reasoning, the velocity profile should depend on the similarity variable:
$\frac{u}{u_0} = \phi\left(\frac{y}{\sqrt{\nu t}}\right)$
> [!important] Similarity Variable
> The characteristic length scale grows as $\sqrt{\nu t}$, reflecting the diffusive nature of momentum transport.
Define the similarity variable:
$\boxed{\eta = \frac{y}{2\sqrt{\nu t}}}$
and the dimensionless velocity:
$\boxed{\frac{u}{u_0} = F(\eta)}$
---
## Transformation to ODE
We now transform the PDE by computing the necessary derivatives.
### Time derivative:
$\frac{\partial u}{\partial t} = u_0 \frac{dF}{d\eta} \frac{\partial \eta}{\partial t}$
where:
$\frac{\partial \eta}{\partial t} = -\frac{1}{4} \frac{y}{\sqrt{\nu} t^{3/2}} = -\frac{\eta}{2t}$
Therefore:
$\frac{\partial u}{\partial t} = u_0 F' \cdot \left(-\frac{\eta}{2t}\right) = \frac{-u_0 F' \eta}{2t}$
### Spatial derivatives:
$\frac{\partial u}{\partial y} = u_0 \frac{dF}{d\eta} \frac{\partial \eta}{\partial y} = u_0 F' \frac{1}{2\sqrt{\nu t}}$
$\frac{\partial^2 u}{\partial y^2} = u_0 F'' \frac{1}{2\sqrt{\nu t}} \frac{\partial \eta}{\partial y} = \frac{u_0 F''}{4\nu t}$
### Substituting into the PDE:
$\frac{-u_0 F' \eta}{2t} = \nu \cdot \frac{u_0 F''}{4\nu t}$
Simplifying:
$\boxed{F'' + 2\eta F' = 0}$
This is an ordinary differential equation for $F(\eta)$.
---
## Transformed Boundary Conditions
The original boundary conditions become:
$\begin{align}
\eta = \frac{y}{2\sqrt{\nu t}} &\quad \Rightarrow \quad \begin{cases}
t = 0: & \eta \to \infty \\
y \to \infty: & \eta \to \infty \\
y = 0: & \eta = 0
\end{cases}
\end{align}$
Therefore:
$\begin{align}
F(\eta \to \infty) &= 0 \\
F(\eta = 0) &= 1
\end{align}$
---
## Solving the ODE
Starting with:
$F'' + 2\eta F' = 0$
Rearrange:
$\frac{dF'}{d\eta} = -2\eta F'$
Separate variables:
$\frac{dF'}{F'} = -2\eta \, d\eta$
Integrate:
$\ln(F') = -\eta^2 + C$
Exponentiate:
$F' = A e^{-\eta^2}$
Integrate again:
$\boxed{F = A \int_0^{\eta} e^{-s^2} \, ds + B}$
---
## Applying Boundary Conditions
At $\eta = 0$: $F(0) = 1$
$B = 1$
As $\eta \to \infty$: $F(\infty) = 0$
$0 = A \int_0^{\infty} e^{-s^2} \, ds + 1$
Using the Gaussian integral:
$\int_0^{\infty} e^{-s^2} \, ds = \frac{\sqrt{\pi}}{2}$
We get:
$0 = A \frac{\sqrt{\pi}}{2} + 1 \quad \Rightarrow \quad A = -\frac{2}{\sqrt{\pi}}$
---
## Final Solution
Substituting the constants:
$\boxed{F(\eta) = 1 - \frac{2}{\sqrt{\pi}} \int_0^{\eta} e^{-s^2} \, ds}$
### Error Function Representation
The error function is defined as:
$\erf(\eta) = \frac{2}{\sqrt{\pi}} \int_0^{\eta} e^{-s^2} \, ds$
Therefore:
$\boxed{F(\eta) = 1 - \erf(\eta)}$
or equivalently, using the complementary error function $\erfc(\eta) = 1 - \erf(\eta)$:
$\boxed{\frac{u}{u_0} = \erfc\left(\frac{y}{2\sqrt{\nu t}}\right) = 1 - \erf\left(\frac{y}{2\sqrt{\nu t}}\right)}$
---
## Physical Interpretation
> [!summary] Key Results
>
> 1. Diffusive spreading: The velocity profile spreads into the fluid with a characteristic length scale $\delta(t) \sim \sqrt{\nu t}$
>
> 2. Self-similar evolution: At different times, the velocity profiles collapse onto a single curve when plotted against $\eta = y/(2\sqrt{\nu t})$
>
> 3. Momentum diffusion: The kinematic viscosity $\nu$ acts as a diffusivity for momentum, exactly analogous to thermal or mass diffusion
>
> 4. Penetration depth: At any given time, most of the velocity change occurs within $y \lesssim \sqrt{\nu t}$
The solution demonstrates that viscosity is fundamentally a diffusive process for momentum transport, validating the kinetic theory picture developed in [[2-What-is-Viscosity|Lecture 2]].
![[StokesFirstProblem.pdf]]